I have two real functions $f(g)$ and $g(x)$.
For every real value $x$, $f(x)$ and $g(x)$ are monotonically increasing with $f(x) > 0$ and $g(x) > 0$.
When the following limits exist:
$$ \lim_{x\to ∞} \dfrac{f(x)}{g(x)} = 1, \lim_{x\to -∞} \dfrac{f(x)}{g(x)} = 1 $$ then, $$ \sup_{-∞ < x <∞}\dfrac{f(x)}{g(x)} < ∞ $$ is established.
Now, I would like to prove that for every $x$, $\dfrac{f(x)}{g(x)} \le c$, the constant $c < ∞$ exists.
It seems that I should start with the epsilon-delta method to prove this. However, I have no idea about constructing a strict and full procedure to do this. I have seen something similar and tried writing as below:
$$ \lim_{x\to ∞}\dfrac{f(x)}{g(x)} = 1, \forall \epsilon_1 > 0, \exists x_1, \\\ |\dfrac{f(x)}{g(x)} - 1| < \epsilon_{1}, 1 - \epsilon_{1} < \dfrac{f(x)}{g(x)} < 1 + \epsilon_{1} \\\ \lim_{x\to -∞}\dfrac{f(x)}{g(x)} = 1, \forall \epsilon_2 > 0, \exists x_2, \\\ |\dfrac{f(x)}{g(x)} - 1| < \epsilon_{2}, 1 - \epsilon_{2} < \dfrac{f(x)}{g(x)} < 1 + \epsilon_{2} $$
$$ x \in (-∞, x_1) \cup (x_2, +∞), \dfrac{f(x)}{g(x)} < max\{1+\epsilon_{1}, 1+\epsilon_{2}\} = k, k \in R \\\ $$
My wondering is that I am not sure how to connect with the value c. I can imagine the two functions on the Cartesian plane, and the change in the x-axis and y-axis. Can I say that k = c in this case?
On $[x_1,x_2]$ we have $\frac {f(x)}{g(x)} \leq \frac {f(x_2)}{g(x_1)}$. Let $c_1= \frac {f(x_2)}{g(x_1)}$ and take $c=\max \{c_1,k\}$.
[The case $x_1 >x_2$ is simpler so I am assuming that $x_1 \leq x_2$].